Question:
When a uniform bar of length \(l\), breadth \(b\) and thickness \(d\) is supported by rigid supports near the ends and loaded at the centre by a vehicle of mass \(M\), the bar sags by an amount \(\delta\). If \(Y\) is the Young's modulus of the material, then \(\delta\) is
When a uniform bar of length \(l\), breadth \(b\) and thickness \(d\) is supported by rigid supports near the ends and loaded at the centre by a vehicle of mass \(M\), the bar sags by an amount \(\delta\). If \(Y\) is the Young's modulus of the material, then \(\delta\) is
Show Hint
For a rectangular beam supported at both ends and loaded at the centre,
\[
\delta=\frac{Wl^3}{4Ybd^3}.
\]
Remember that the depression is directly proportional to the load and cube of the length, and inversely proportional to \(Y\) and \(d^3\).
Updated On: Jul 29, 2026
- \[ \frac{Ml^3}{4bd^3Y} \]
- \[ \frac{Yl^3}{4Mgbd^3} \]
- \[ \frac{Yd^3}{4Mgl^3b} \]
- \[ \frac{Mgl^3}{4bd^3Y} \]
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For a beam supported at both ends and loaded at the centre, the depression (sag) at the centre is given by
\[
\delta=\frac{Wl^3}{4Ybd^3},
\]
where
\[
W=\text{load},
\qquad
Y=\text{Young's modulus}.
\]
Step 1: Determine the load acting on the bar. The vehicle of mass \(M\) exerts a force \[ W=Mg. \]
Step 2: Substitute into the bending formula. Using \[ \delta=\frac{Wl^3}{4Ybd^3}, \] we get \[ \delta = \frac{(Mg)l^3}{4Ybd^3}. \] \[ \delta = \frac{Mgl^3}{4bd^3Y}. \]
Step 3: Identify the correct option. Hence, \[ \boxed{ \delta= \frac{Mgl^3}{4bd^3Y} } \] Therefore, \[ \boxed{\text{Answer = (D)}} \]
Step 1: Determine the load acting on the bar. The vehicle of mass \(M\) exerts a force \[ W=Mg. \]
Step 2: Substitute into the bending formula. Using \[ \delta=\frac{Wl^3}{4Ybd^3}, \] we get \[ \delta = \frac{(Mg)l^3}{4Ybd^3}. \] \[ \delta = \frac{Mgl^3}{4bd^3Y}. \]
Step 3: Identify the correct option. Hence, \[ \boxed{ \delta= \frac{Mgl^3}{4bd^3Y} } \] Therefore, \[ \boxed{\text{Answer = (D)}} \]
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